Compute probability using indicator function I am proofing the Tail Probability Expectation Formula, I have some idea but I don't know it is correct or not.
$$ E[X] = \sum_{n=1}^{\infty}P\{X \geq n\}$$
$$$$
First I create an indicator function.
$$ 1_{\{X>n\}}=
 \begin{cases} 
      1 & ,x>n \\
      0 & ,otherwise 
   \end{cases}
$$
Then,
$$E[X] = E[E[X|1_{\{X>n\}}]]$$
$$ E[X] = E[P\{X>n\}]$$
Since X is non-negative, we have $n>0$ (I am wondering it is correct or not). We have...
$$E[X] = \sum_{n=0}^{\infty}P\{X>n\}$$
Then increment the index by 1 since $n>0$.
$$ E[X] = \sum_{n=1}^{\infty}P\{X \geq n\}$$
Hence
as desired.
 A: Hmm, I don't see how you go from your first $E(X)$ to your second... $P(X > n) = E(1_{X > n})$.
Here's a hint for how to do it. Recall that
$$ E(X) = \sum_{n=1}^\infty n P(X = n). $$
So you have one lot of $P(X = 1)$, two of $P(X = 2)$, three of $P(X = 3)$, etc. Imagine these 'boxes'. Each number $n$ represents $P(X = n)$
123456789...
 23456789...
  3456789...
   456789...
    56789...
 ... ... ...

There is one 1 because you count $P(X = 1)$ once; there are two 2s because you count $P(X = 2)$ twice. If you add up all of these, you get $E(X)$. The formula
$$ E(X) = \sum_{n=1}^\infty n P(X = n) $$
corresponds to adding up column by column: first $n = 1$ does the one term in te first column, then $n = 2$ does the two terms in the second column, etc.
Instead, try to add up all the elements a row at a time. Your first term is the entire first row:
$$ P(X = 1) + P(X = 2) + P(X = 3) + \cdots = P(X \ge 1). $$
Your second term is the second row:
$$ P(X = 2) + P(X = 3) + P(X = 4) + \cdots = P(X \ge 2). $$
Can you show rigorously that the $n$-th term is $P(X \ge n)$? If so, once you sum up all the rows, you can the desired formula
A: The equation $E[X]=E[E[X\mid 1_{X\ge n}]]$ does not imply $E[X]=E[P\{X>n\}]$. However, here is a method possibly more in line with your original idea.
We start with the following deterministic equality, valid for any nonnegative integer $x$:
$$
x=\sum_{n=1}^{\infty} {\bf 1}_{x\ge n}
$$
In words, a nonengative integer is equal to the number of positive integers which are less than or equal to it.
Now, replace $x$ with the random variable $X$ to get $X=\sum_{n=1}^\infty 1_{X\ge n}$, and take the expected value of each side. Done!
